= Solution
Assume zero temperature, complete ionization, constant composition, noninteracting electrons, negligible ion thermal pressure, and Newtonian stellar gravity. Ions supply almost all the mass, while <electron degeneracy pressure> supplies support. By the <Pauli exclusion principle>, the two electron spin states fill a momentum sphere up to <Fermi momentum> $p_F$. Counting states gives
$$
n_e=\frac{2}{(2\pi\hbar)^3}\frac{4\pi p_F^3}{3}
=\frac{p_F^3}{3\pi^2\hbar^3},\qquad
\rho=\mu_em_un_e,
$$
where $\mu_e$ is <mean molecular weight per electron>. The momentum flux of this isotropic <Fermi gas> is
$$
P=\frac{2}{3(2\pi\hbar)^3}\int_{p\le p_F}p\,v(p)\,d^3p
=\frac1{3\pi^2\hbar^3}\int_0^{p_F}\frac{p^4c^2}{\sqrt{m_e^2c^4+p^2c^2}}\,dp.
$$
Writing $x=p_F/(m_ec)$ and performing the integral gives the <equation of state of a cold electron gas>
$$
\boxed{P=\frac{m_e^4c^5}{24\pi^2\hbar^3}
\left[x(2x^2-3)\sqrt{1+x^2}+3\operatorname{arsinh}x\right],\qquad
x=\frac{\hbar}{m_ec}\left(\frac{3\pi^2\rho}{\mu_em_u}\right)^{1/3}.}
$$
In the nonrelativistic limit $p_F\ll m_ec$, $v(p)\simeq p/m_e$, so
$$
\boxed{P=K_{\rm NR}\rho^{5/3},\qquad
K_{\rm NR}=\frac{\hbar^2(3\pi^2)^{2/3}}{5m_e(\mu_em_u)^{5/3}}.}
$$
In the ultrarelativistic limit $p_F\gg m_ec$, $v(p)\simeq c$, so
$$
\boxed{P=K_{\rm R}\rho^{4/3},\qquad
K_{\rm R}=\frac{\hbar c(3\pi^2)^{1/3}}{4(\mu_em_u)^{4/3}}.}
$$
These are <polytropic equations of state> with indices $3/2$ and $3$, respectively. The relativistic softening underlies the <Chandrasekhar mass> limit; Coulomb corrections, thermal effects, rotation, and general relativity are excluded from this idealized derivation.
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